dr. sangeeta khanna ph.d 1 · a chloro compound (x) on treatment with zn/hcl gave a hydrocarbon...
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Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+1\Org\Test\GT-7\Grand Test -7 (organic Chemistry) Level -2.docx
Dr. Sangeeta Khanna Ph.D
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LEVEL – II
TOPIC: HYDROCARBON & GENERAL ORGANIC CHEMISTRY
READ INSTRUCTIONS CAREFULLY
1. The test is of 1 hour duration.
2. The maximum marks are 161.
3. This test consists of 37 questions.
SECTION – A (Single Answer Type) Negative Marking [-1]
This Section contains 15 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only One choice) 15 × 4 = 60 Marks
1. Cyclohexene when treacted with Br2 in CCl4 gives:
a. Bromocyclohexane b. trans –1, 2 – Dibromocyclohexane
c. cis – 1, 2 – Dibromocyclohexane d. 1, 1-Dibromocyclohexane
B
Sol. Addition of Br2 is Anti-addition; Br2 will add on opposite side to ring.
2. The treatment of benzene with isobutene in the presence of sulphuric acid gives:
a. Isobutyl benzene b. ter–Butyl benzene
c. n–Butyl benzene d. No reaction
B
Sol.
3. A chloro compound (X) on treatment with Zn/HCl gave a hydrocarbon with five carbon atoms in the
molecule. When (X) is dissolved in ether and treated with sodium 2,2,5,5-tetramethylhexane is
obtained. The compound (X) is
a. 2-Chloro-2-methylpropane b. 1-Chloro-2, 2-dimethylpropane
c. 2-Chloro-2-methylbutane d. Isopropyl chloride
B
Sol. 3
|
|22
|
|3
actionReWurtz2
|
|3 CH
CH
CH
CCHCH
CH
CH
CCHClCH
CH
CH
CCH
3
3
3
3
3
3
4. The reaction of one equivalent of HBr with CH2 = CH – C CH gives
a. CH2 = CH – C CBr b. CH2 = CH – CBr = CH2
c. CH3 – CHBr – C CH d. CH2 = CH – CH = CHBr
B Sol. First addition on triple bond due to formation of stable product a conjugated diene.
+ H3C – C = CH2
CH3
H+ C CH3
CH3
CH3
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5. The bond dissociation energy of the C – H bond for the compound
)P(
HCH3
)Q(
HCHCH 23
decreases in the order: a. P > Q > R > S b. S > R > Q > P c. S > P > Q > R d. Q > P > S > R C Sol. Bond dissociation energy can be compared by stability of free radical
CH3
radical benzy lstabilise
Resonance2
gationhy perconjuby Stabilise
23 HCPh HCCH
6. :oductPr)major(
Mole) (1 H2
a. b. c. d. None of these B Sol. 7. Which of the following molecules is expected to have the greatest resonance stabilization?
a. b. c. d. B Sol.
8. Consider the following reaction sequence In this B and D respectively are a. b.
Highly destabilise
4n + 2 = 6 Aromatic
4n + 2 = 6 Aromatic
Cl
NO2
and
Cl
NO2
Cl
NO2
and NO2
Cl
Cl2/FeCl3
HNO3/H2SO4
A HNO3/H2SO4
B
C Cl2/FeCl3
D Below 330K
H
(S)
CH2 – H
(R)
H
H
1, 4-addition product is major
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c. d. A Sol. Remember : – Cl is o, p-directing while –NO2 is meta directing group. 9. a. R > S > Q > P b. P > S > R > Q c. Q > P >S > R d. S > R > Q > P D Sol. As - H decrease, stability of anion increases.
Stability of carbanion
)gationHy perconju(
effect donating e
1
10. Which of the following acids has lowest pKa value?
a. OH
O
CH
OH
C||
2|
b. OH
O
CCHH
NO
C| |
22
2
|
c. OH
O
CH
Cl
C| |
2|
d. OH
O
C
F
F
CCH|||
|3
D
Cl
SO3H
and
Cl
NO2
Cl
NO2
and
NO2
Cl NO2
CH2
CH3
(P)
CH2
CH2CH3
CH2
CH CH3
CH3
CH2
CH3
H3C – C – CH3
(Q) (R) (S)
Cl2/FeCl3 HNO3 Below 330K
Cl NO2
HNO3/H2SO4 Cl2/FeCl3
Cl NO2
NO2
Cl
o & p-directing (meta-directing)
o –isomer +
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Sol. ,COOH
F
F
CCH|
|3 2 (–I) groups ; is most acidic & have lowest pKa & highest Ka
11. In given structure which lone pair is localised a. 1 & 2 b. 2 & 4 c. 1 & 4 d. 2 & 3 C
Sol. Lone pair on doubly bonded atom will not be delocalized. 12. Which of the most preferred site of electrophilic attack in the following compound.
a. 2 & 3 b. 6 & 4 c. 1 & 3 d. 1, 3 & 5 B
Sol.
HN with lone pair will exert e– donating effect & makes the ring activating at o & p-position
13. Which Benzene ring is most reactive in electrophilic attack a. (1) b. (2) c. (3) (d) Same Reactivity B
Sol.
(+ve charge is delocalized in both the ring. More structure have aromatic ring in carbocation)
14. An alkyl bromide (A) forms Grignard’s reagent which on treatment with water yields n-hexane . (A) with sodium in presence of dry ether forms 4,5-diethyloctane. (A) is:
a. CH3[CH2]5Br b. CH3[CH2]3CH(Br)CH3 c. CH3 – [CH2]2 – CH(Br)CH2CH3 d. CH3[CH2]2CH(Br) CH = CH2
C
NH O
1
3
2
4
5
6
1 2 3
E
N
H NH
N
NH H
1
2
4
3
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Sol. 15. Treatment of 2-methyl-2-butene with HBr in the presence of a peroxide yields
a. A primary alkyl bromide b. A secondary alkyl bromide c. A tertiary alkyl bromide d. A vicinal dibromide B
Sol. 3|
|
|3
H
Radical f ree in entrearrangem no
3
||
3Peroxide
HBr3
|
3 CHH
Br
C
CH
H
CCHCHH
Br
C
CH
CCHCHCH
CH
CCH
333
SECTION – C (Comprehension Type) Negative Marking [-1]
This Section contains 2 Comprehension. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 9 × 4 = 36 Marks
Passage 1
When (C – H) sigma electrons are in conjugation with pi bond, this conjugation is knwon as (C – H)
conjugation, excess conjugation or hyperconjugation. (i) Compound should have at least one sp2-hybrid carbon of either alkene, alkyl carbocation or alkyl
free radical.
(ii) -carbon with respect to sp2 hybrid carbon should have at least one hydrogen. (iii) Resonating structures due to hyperconjugation may be written involving "no bond” between the
alpha carbon and hydrogen atoms
2
Θ|
2
Θ|
|2
Θ|
|2
|
|HCCH
H
H
CHHCCH
H
H
C HHCCH
H
H
CHCHCH
H
H
CH
In the above resonating structures there is no covalent bond between carbon and hydrogen, and from this pooint of view, hyperconjugation may be regarded as "no bond resonance”. Actually the hydrogen atom is not free from the carbon. These resonating structures only suggest that: (a) there is some ionic character between C – H bond and (b) carbon-carbon double bond acquires some single bond character. We can explain the stability of alkene, carbocation and carbon free radical on the basis of hyperconjugation.
ionhydrogenat of Heat
1H- of number alkene of Stability
CH3 – CH2 – CH2 – CH – CH2CH3
Br
Mg CH3CH2CH2 – CH – CH2 – CH3
MgBr
HOH Hexane
Na + dry ether
CH3 – CH2 – CH2 – CH – CH – CH2 – CH2 – CH3
CH2
CH3
CH2
CH3
4,5-Diethyloctane
Dr. Sangeeta Khanna Ph.D
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1. Which of the following statements are correct for C6H5 – CCl3? (a) CCl3 group is electron withdrawing due to the –I effect and reverse hyerpconjugation. (b) CCl3 group is meta directing due to the – M effect. (c) CCl3 group is o, p-directing because it is +R group. (d) CCl3 group can exert +M effect. A
2. Which of the following has the lowest heat of hydrogenation?
(a) (b) (c) (d)
D Sol. Maximum Hyperconjugation. Means least stable alkene 3. Carbon-carbon double bond length will be maximum in which of the following compounds?
(a) CH3 – CH = CH2 (b) CH3 – CH = CH – CH3
(c) 3
|
|3 CH
CH
C
CH
CCH
3
3
(d) CH2 = CH2
C
Sol. Maximum hyperconjugation will make – C = C –; a single bond
4. The heats of combustion for the four C6H12 isomers shown are (not necessarily in order): 955.3, 953.6, 950.6, and 949.7 (all in kilocalories per mole). Which of these values is most likely the heat of combustion of isomer 1?
a. 955.3 kcal/mol b. 950.6 kcal/mol c. 953.6 kcal/mole d. 949.7 kcal/mol
B Sol. Stable alkene have lowest heat of Hydrogenation. Stability order is 2 > 1 > 3 > 4; 2nd lowest value
for 1 5. The heats of combustion for the four C6H12 isomers shown are (not necessarily in order): 955.3,
953.6, 950.6, and 949.7 (all in kilocalories per mole). What can be said about isomers 3 and 4?
a. 3 is more stable by 1.7 kcal/mol. b. 3 more stable by 3.0 kcal/mol. c. 4 is more stable by 1.7 kcal/mol. d. 3 is more stable by 0.9 kcal/mol. A
Sol. )3(6.950 ;
)4(7.949 ; difference is 1.7 kcal/mol. Isomer 3 is more stable by 1.7.
1 2 3 4
1 2 3 4
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Passage – 2
6. Which of the following is compound P? (a) (b) (c) (d) A
Sol. As this compound on dehydrohalogenation will form non terminal alkyne Which don’t react with Na. 7. R and S are:
(a) position isomers (b) enantiomers (c) geometrical isomers (d) functional group isomers C
Sol. R = cis-pent-2-ene; S = trans pent-2-ene 8. Identify structure of compound T:
(a) OH
O
CCH||
3 (b)
(c) OH
O
CCHCH||
23 (d) H
O
CCH||
3
B Sol. Alkyne on ozonolysis form dione;
2|
||||
3OH/Zn .2
O .1323 H
CH
C
O
C
O
CCHCHCHCCCH
3
2
3
O
O
Br
Br Br
P(C5H9Br)
NaNH2
Q Decolourise Br2 water
and cold KMnO4 Q T 1. O3
2. Zn/H2O (No reaction with Na)
Lindlar catalyst Na/NH3()
R (C5H10)
S (C5H10)
(1) O3
(2) (CH3)3S Z + Y
Br
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9. Compound z & y are a. CH3CHO + CH3CH2CHO b. CH3COOH + CH3CH2COOH
c. CHOCHCHOH
CH
CCH 3
|
3
3
d. Pentane-2, 3-dione
A
Sol. OS\/
CH
CHCHOCHCHCHOCHCHCHCHCHCH
3
3
233Ozonoly sis
323
SECTION – D (More than One Answer Type) No Negative Marking
This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 5 × 5 = 25 Marks
1. Which of the following alkenes are more stable than ? a. b. c. d. A,B,C,D Sol. A = 10 CH; B = 5 – CH; C = 4CH; D = 4 – CH; given compound have – 3CH 2. Which of the following compound on oxidation with acidic KMnO4 will give Benzoic acid?
a. Ethylbenzene b. Benzaldehyde c. Phenylmethanol d. Toluene A,B,C,D
Sol.
3. )X(CH
H
C
H
CCH 2Br
)cis(
3||
3 ;
The product (X) is:
a. formed by anti addition of Br2 b. racemic mixture of ‘d’ and ‘’ form c. CH3CHBrCHBrCH3 d. formed by syn addition of Br2 A,B,C
CH2CH3 COOH CH3OH
CH3
CHO
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4. Which of the following compound on Reaction with Red P + HI will give 2-Methyl propane a. 2-Methylpropan-2-ol b. 2-methylbutanoic acid c. Isobutyl iodide d. 2-methyl propanal A,C,D
Sol. (A) 3
|
3H P dRe
3
|
|3 CHH
CH
CCHCH
CH
OH
CCH
33
I
(B) etanMethy lbu2
32
|
3H P dRe
2
|
3 CHCHH
CH
CCHCOOHCHH
CH
CCH
33
I
(C) 3
|
3HIP Red
2
|
3 CHH
CH
CCHCHH
CH
CCH
33
I
(D) 3
|
3HIP dRe
|
3 CHH
CH
CCHCHOH
CH
CCH
33
5. Ni/H2
33CH
DCC
CH
D
/\
\
/
Product of above reaction will be:
a. racemic mixture b. optically inactive c. meso d. constitutional isomers B,C
Sol.
SECTION – E (Matrix Type) No Negative Marking
This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. 8 × 2 = 16 Marks
1. Column (I) Column (II)
(a) 23 HCCH
(p) Mesomeric effect
(b) H3C - 2HC
(q) Hyperconjugation
(c) Cl
Cl
CClΘ
| (r) + I effect
(d) (s) Reverse Hyperconjugation
Sol. A Q, R; B Q, R; C S; D P, R 2. Match Column (I) with Column (II)
(a) (p) Nucleophilic substitution
C O
O
H Θ
+ Ph – CH2 – Cl AlCl3
CH2 – Ph
H3C
D C = C
D
CH3 H2/Ni H3C
D C C
D
CH3
H H
Meso (Plane of symmetry) Syn addition
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(b) (q) Electrophilic substitution (c) (r) Cation intermediate (d) (s) Free radical substitution
Sol. A – (q), (r); B – (q), (r) ; C – (p); D – (s)
SECTION – F (Integer Type) No Negative Marking
This Section contains 6 questions. The answer to each question is a Single Digit Integer ranging from 0 to 9. (6 × 4 = 24 Marks)
1. Examine the structural formulas of following compounds and find how many compounds will produce
CO2 on with hot acidic KMnO4 (a) , (b) , (c) , (d)
(e) , (f) H3C – C CH, (g) Ph – CH = CH2
Sol. 5 (b); (c); (e); (f); (g) (terminal alkene & alkyne) 2. How many carbocations undergo rearrangements?
(i) (ii) (iii) 3
|
23 CH
CH
CCHCH
3
(iv) (v) (vi) (vii) (viii) (ix) (x)
Sol. 6; (i), (ii), (iv), (v), (vi); (x) (i) methylshift (ii) ring contraction, (iv) & (v) ring expansion; (vi) H– shift (x) H– shift
H3C – C – CH3
CH3
CH3
Br2, hv CH3 – C – CH2 – Br
CH3
CH3
CH2
Me3C – CH2 – CH2
CH2
CH2 = CH
CH2 = CH – CH2
+ HNO3 H2SO4 NO2
Br
KCN CH3 – CH – CH3
CN
+ KBr
HC – CMe2 Ph
Ph
C – CH3
OH
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3. From the following compounds/ ions:
(a) 3HC
(b) 4HN
(c) BF3 (d) NH3 (e) AlCl3 (f) F– (g) 2CCl (h) CH2 = CH2
Identify value of X. Where X is the total number of electrophiles. Sol.4 (a, c, g, e) 4. How many of the following halide on reaction with Mg, dry ether followed by acidification will form 2-
methyl pentane
(a) 32
||
3 CHCHH
Br
CH
CH
CCH
3
(b) (c)
(d) BrH
CH
CCH)CH(
3|
23 (e) 3|
|
23 CHH
Br
CH
CH
CCHCH
3
(f) 3||
|
3 CHH
CH
CH
Br
CH
CH
CCH
3
3
(g) (CH3)3C – CH2Br Sol. 3 (a), (b), (c)
2|
2
|
3H
32
||
3Mg
32
||
3 H
CH
CCH
CH
CCHCHCHH
MgBr
CH
CH
CCHCHCHH
Br
CH
CH
CCH
3
333
5. How many of following reactions are electrophilic addition
(a) 3|
3H
HOH23 CHH
OH
CCHCHCHCH
(b) OH
CH
CH
CCHH
CH
CH
CCH
3
3
4
3
3
|
|3
H
KMnO|
|3
(c) (d)
(e) 3
|
|3
HCl2
|
3 CH
CH
Cl
CCHCH
CH
CCH
33
(f) 3
||||
3Ozonoly sis
33 CH
O
C
O
CCHCHCCCH
Ans. 2 (a, e) (B = oxidation (free Radical); C = free Radical addition; D = electrophilic subsitution 6. How many trichloroethanes would be produced when 1,1-Dichloroethane reacts with chlorine ? Sol. 2;
H3C – CH – CH2 – CH2
CH2 – Br
CH3
+ CH3Cl Anhyd.
AlCl3 + HCl +
CH3
CH3 – C – CH2
CH3
Br
CH2
CH3
+ Cl2 h
Cl
Cl
Cl
Cl
Cl
Cl
CH3 – C – Cl
Cl
Cl
& CH3 – CH Cl
Cl
Cl2 h
CH3 – CH Cl
Cl
Cl